Combining two ratios of the same mass to create a new ratio

Question

This was stuck on this question wondering if you could help?

Two solutions, each of mass 1L, has a ratio of liquid A to liquid B as 1:6 and 1:8 respectively. Both solutions are mixed together to create one

Find the new ratio of liquid A to liquid B in the new solution

I would also like to know how this could be done for any other values in weight or ratio

My tried answer:

Liquid A in first sol= 1/7 * 1L = 1/7 L

Liquid A in second sol= 1/9 * 1L = 1/9 L

Liquid A new sol = 1/7 + 1/9 = 16/63 L

Ratio = (16/63):(110/63) = 16:110 = 8:55

Answer 1

$|A|=\frac{1}{7}$ and $|B|=\frac{6}{7}$ in the first solution.

$|A|=\frac{1}{9}$ and $|B|=\frac{8}{9}$ in the second solution.

Thus, the needed ratio it's:$$\frac{\frac{1}{7}+\frac{1}{9}}{\frac{6}{7}+\frac{8}{9}}=8:55$$

Answer 2

Suppose you have $n$ liquids with masses $V_1,V_2,...,V_n$ and ratios of $A:B$ as $a_1:b_1,a_2:b_2,...,a_n:b_n$ respectively.

For each liquid, the mass of $A$ is $V^A_i=\frac{a_i}{a_i+b_i}V_i$ and the mass of $B$ is $V_i^B=\frac{b_i}{a_i+b_i}V_i$. Note that $V_i^A+V_i^B=V_i$ for all $i$, as expected.

Now you combine them into a homogeneous liquid of mass $V=\sum_{i=1}^nV_i$. This combined liquid has mass of $A,V^A=\sum_{i=1}^nV_i^A=\sum_{i=1}^n\frac{a_i}{a_i+b_i}V_i$ and mass of $B,V^B=\sum_{i=1}^nV_i^B=\sum_{i=1}^n\frac{b_i}{a_i+b_i}V_i$. Once again note that $V^A+V^B=V$ as expected.

Thus $$\frac{V^A}{V^B}=\frac{\sum_{i=1}^n\frac{a_i}{a_i+b_i}V_i}{\sum_{i=1}^n\frac{b_i}{a_i+b_i}V_i}=\frac{V^A}{V-V^A}$$Can you apply this scheme to your question and find the answer?